aekotra
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Reactivity of solutions with the same anion
Hi,
Maybe you guys could help me with a question: If I have two aqueous solutions of two different compounds (with different solubility) that result in
equal amounts of the same anion, should the reactivity of that anion in the solutions be different or the same?
The specific example:
Solution A
CaF2 (aq)
Solubility: 0.0016 g/100 mL (20 °C)
Solution B
Na2SiF6 (aq)
Solubility: 0.64 g/100 mL (20 °C)
SiF<sub>6</sub><sup>2-</sup> + 2 H<sub>2</sub>O → 6 F<sup>-</sup> +
SiO<sub>2</sub> + 4 H<sup>+</sup>
The solutions are prepared such that both contain the same amount of F-. Is the reactivity of F- in both solutions the same or
different?
Annoying typo removed!
[Edited on 17-1-2013 by ScienceSquirrel]
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woelen
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The solutions are very different. In the second solution, there will hardly be any free F(-) ions. The complex ion SiF6(2-) is quite stable and
remains as such in aqueous solution.
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aekotra
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Quote: Originally posted by woelen  | | The solutions are very different. In the second solution, there will hardly be any free F(-) ions. The complex ion SiF6(2-) is quite stable and
remains as such in aqueous solution. |
This seems not to be the case as Na2SiF6 is used to fluoridate drinking water and is produced from hexafluorosilicic acid. It 'hydrolyzes rapidly' apparently.
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woelen
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This rapid hydrolysis apparenty only occurs at very low concentrations, I have ammonium hexafluoro silicate and if I dissolve this in water, then
nearly 100% remains present as hexafluorosilicate ion. I can imagine that very dilute solutions have a much larger percentage of hydrolysis. Still, I
think that the available fluoride ions in solutions are lower than if the same amount of fluorine atoms were available from e.g. NaF. But if you want
more precise info, then you'll probably need to find the equilibrium constants for the hydrolysis reaction. I hardly can imagine that these have not
been investigated, especially because of the widespread use of hexafluorosilicates for fluoridation of water.
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ScienceSquirrel
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Fluoride is a bad choice due to it's tendency to form hydrogen bonds, complexes etc.
If you choose alkali metal salts of a simple anion eg sodium bromide and potassium bromide then a dilute solution of either salt will behave as a
simple source of bromide anions regardless of the cation.
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